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kirill115 [55]
3 years ago
9

An attempt at synthesizing a certain optically active compound resulted in a mixture of its enantiomers. The mixture had an obse

rved specific rotation of 14.1°. If it is known that the specific rotation of the R enantiomer is –28.4°, determine the percentage of each isomer in the mixture. g
Chemistry
1 answer:
Lemur [1.5K]3 years ago
8 0

Answer:

The percentage of the R-enantiomer is 26.18% while the percentage of the S-enantiomer is 73.82%

Explanation:

If the specific rotation of R enantiomer = -28.4, then the specific rotation of S = +28.4

Now, let us have x = % R, thus

% S = 100-x =y

Hence;

{- 28.4x + 28.4( 100 -x)}/100= 14.1

Thus;

-28.4x + 2840 -28.4x = 1410

-56.8x + 2840 = 1410

-56.8x = 1410-2840

-56.8x = -1430

x = 1430/56.8

x = 26.18%

y = 100-26.18% = 73.82%

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